A solid sphere of radius 10 cm is rotating about an axis which is at a distance 15cm from its centre. The radius of gyration about this axis is$$\sqrt{n}cm$$. The value of n is

We need to find n where the radius of gyration about the axis at 15 cm from center is $$\sqrt{n}$$ cm.

Solid sphere, R = 10 cm, axis at distance d = 15 cm from center.

Using parallel axis theorem:

$$I = I_{cm} + Md^2 = \frac{2}{5}MR^2 + Md^2$$

Radius of gyration: $$k^2 = \frac{I}{M} = \frac{2}{5}R^2 + d^2 = \frac{2}{5}(100) + 225 = 40 + 225 = 265$$

$$k = \sqrt{265}$$ cm, so n = 265.

Therefore, n = 265.